34. If R = 4.0 kΩ, C = 3.0 mF, ε = 15 V, Q = 12 mC, and I = 2.0 mA, what is the potential difference Vb − Va?

the potential difference Vb − Va = -3 V

Explanation:

The potential difference Vb − Va s given by

Vb − Va=  Capacitance / Charge + current x resistance - Electromotive force

Vb − Va=Q/C + I X R - E

V=Q / C=(12 x 10^-3) / (3 x 10^-3)

=0.012/0.003ss

=4V

alsoV=  1R = 2.0 x 10^-3 x 4 x 10^3

=0.002 X 4000

=8

Vb − Va= 4V + 8V- 15V

= -3V

Therefore, the potential difference Vb − Va = -3 V

the order for strong to weak forces such as

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

the fundamental forces are those forces that act between to objects or element to make interactions. there are four types of fundamental forces such as strong and weak nuclear forces (due to protons and electrons), electromagnetic force (due to the motion of electrons) and gravitational force (due to orbital movement of the planet). if these forces are arranged on the basis of their maximum strength then the order is

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down